Problem: Let $a(x)=4x^6+5x^5-6x^2+2x+1$, and $b(x)=x^3$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Answer: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's rewrite the fraction to cancel common factors: $ \begin{aligned} \dfrac{a(x)}{b(x)}=\dfrac{4x^6+5x^5-6x^2+2x+1}{x^3}&=\dfrac{4 {x^6}+5 {x^5}}{ {x^3}}+\dfrac{-6x^2+2x+1}{x^3}\\\\ &={4x^3+5x^2}+\dfrac{{-6x^2+2x+1}}{x^3}\\\\ &={q(x)} + \dfrac{{r(x)}}{b(x)}\end{aligned}$ Since the degree of ${-6x^2+2x+1}$ is less than the degree of $x^3$, it follows that ${r(x)}={-6x^2+2x+1}$, and ${q(x)}={4x^3+5x^2}$. To conclude, $q(x)=4x^3+5x^2$ $r(x)=-6x^2+2x+1$ [Is there another way of doing this?]